A TREATISE OF Angular Sections By JOHN WALLIS D. D. Professor of Geometry in the University of Oxford and a Member of the Royal Society LONDON LONDON Printed by Iohn Playford for Richard Davis Bookseller in the University of OXFORD 1684. A TREATISE OF Angular Sections CHAP. 1. Of the Duplication and Bisection of an ARCH or ANGLE I. LET the Chord or Subtense of an Arch proposed be called A or E of the Double B of the Treble C of the Quadruple D of the Quintruple F c. The Radius R the Diameter 2R But sometimes we shall give the name of the Subtense A E c. to the Arch whose Subtense it is yet with that care as not to be liable to a mistake II. Where the Subtense of an Arch is A let the Versed sine be V where that is E let this be U. Which drawn into or Multiplied by the remainder of the Diameter 2 R − V makes 2 R VVq the Square of the Right-sine this Sine being a Mean-proportional between the Segments of the Diameter on which it stands erect by 13 ◠6. That is Q ½ B the Square of the Right-sine or half the Subtense of the double Arch That is 2 R V − Vq = Q ½ B = ¼ Bq. III. If to this we add Vq the Square of the Versed-sine it makes 2 RV = ¼ Bq + Vq = Aq. And by the same reason 2 R U = Eq. That is IV. The Subtense of an Arch is a Mean Proportional between the Diameter and the Versed-sine V. Again because 2 R V = Aq therefore dividing both by 2 R And the Square thereof Which subtracted from Aq leaves the Square of the Right-sine And in like manner and and That is VI. If from the Square of the Subtense we take its Biquadrate divided by the Square of the Diameter the Remainder is equal to the Square of the Right-sine And the Square-root of that Remainder to the Sine it self And the double of this to the Subtense of the double Arch. VII Accordingly because therefore its Quadruple and And in like manner That is VIII If from Four-times the Square of a Subtense are taken its Biquadrate divided by the Square of the Radius the Remainder is the Square of the Subtense of the double Arch And the Quadratick Root of that Remainder is the Subtense it self IX But That is X. The Rect-angle of the Subtense of an Arch and of its Remainder to a Semicircle divided by the Radius is equal to the Subtense of the double Arch. XI Because therefore AE = RB = 2R × ½B And R. A B. B And therefore because A E contain a Right-angle as being an Angle in a Semicircle XII In a Right-angled Triangle the Rect-angle of the two Legs containing the Right-angle is equal to that of the Hypothenuse and the Perpendicular from the Right-angle thereupon And XIII As the Radius to the Subtense of an Arch so the Subtense of its Remainder to a Semicircle is to that of the double Arch. XIV Because B the Subtense of a double Arch doth indifferently subtend the two Segments which compleat the whole Circumference and consequently the half of either may be the single Arch of this double It is therefore necessary that this Equation have two Affirmative Roots the greater of which we will call A and the lesser E And therefore That is XV. Any Arch and its Remainder to a Semicircumference as also its excess above a Semicircumference and either of them increased by one or more Semicircumferences will have the same Subtense of the double Arch. For in all these Cases the Subtense of the single Arch will be either A or E. XVI Because And therefore and 4Aq Rq − Aqq = Bq Rq = 4Eq Rq − Eqq Therefore by Transposition 4Aq Rq − 4Eq Rq = Aqq − Eqq and dividing both by That is XVII The Square of the Diameter is equal to the difference of the Biquadrates of the Subtenses of two Arches which together complete a Semicircumference divided by the difference of their Squares And this also equal to the sum of the Squares of those Subtenses That is because A E contain a Right-angle XVIII In a Right-angled Triangle the Square of the Hypothenuse 4Rq is equal to the Squares of the sides containing the Right-angle Aq+Eq XIX Or thus Because B is the common Subtense to two Segments which together complete the whole Circumference and therefore the half of both complete the Semicircumference If therefore in a Circle according to Ptolemy's Lemma a Trapezium be inscribed whose opposite sides are A A and E E The Diagonals will be Diameters that is 2R And consequently 4Rq = Aq + Eq as before XX. Hence therefore The Radius R with the Subtense of an Arch A or E being given we have thence the Subtense of the double Arch B which is the Duplication of an Arch or Angle For R A being given we have or R E being given we have A = 4Rq − Eq And having R A E we have by § 9. XXI The Radius R with B the Subtense of the double Arch being given we have thence the Subtense of the single Arch A or E. which is the Bisection of an Arch or Angle For by § 14 And therefore 4Rq Aq − Aqq = Rq Bq = 4Rq Eq − Eqq. And the Roots of this Equation or Eq. And the Quadratick Root of this is A or E. XXII Hence also we have an easie Method for a Geometrical Construction for the Resolution of such Biquadratick Equations or Quadratick Equations of a Plain Root wherein the Highest Power is Negative Understand it in Mr. Oughtred's Language Who puts the Absolute Quantity Affirmative and by it self and the rest of the Equation all on the other side Suppose Rq Bq = 4Rq Aq − Aqq or putting P = ½ B 4Rq Pq = 4Rq Aq − Aqq. For dividing the Absolute term Rq Bq or 4Rq Pq by the Co-efficient of the middle term 4Rq the Result is ¼Bq or Pq and its Root ½ B or P. Which being set Perpendicular on a Diameter equal to 2 R the Square Root of that Co-efficient a streight Line from the top of it Parallel to that Diameter will if the Equation be not impossible cut the Circle or at least touch it From which Point of Section or Contact two streight-streight-lines drawn to the ends of the Diameter are A and E the two Roots of that ambiguous Biquadratick Equation or if we call it a Quadratick of a Plain-root the Root of the Plain-root of such Quadratick Equation XXIII And this Construction is the same with the Resolution of this Problem In a Right-angled Triangle the Hypothenuse being given and a Perpendicular from the Right-angle thereupon to find the other sides and if need be the Angles the Segments of the Hypothenuse and the Area of the Triangle ½ R B or P R. XXIV Or thus Having R and B as at § 22. with the Radius R describe
which is ever equal to χ+δ±μ however these parts be intermingled Which where it is +μ is commonly more obvious to the Eye but where it is − μ is more perplex and will need more consideration to discern but it is equally true in both cases The Square of the Base of an Angle of 135 Degrees is equal to the Squares of the Legs with a Rectangle of them Multiplied into VI. If A be 45 Degrees It will in like manner be shewed that because of B = χ+δ − μ. into χ+δ − μ = B = Bq. That is The Square of the Base of an Angle of 45 Degrees is equal to the Squares of the Legs wanting a Rectangle of them Multiplied into VII And universally what ever be the Angle A it will by like process be shewed That That is The Square of the Base whatever be the Angle at the Vertex is equal to the Squares of the Legs together with if it be greater than a Right-angle or wanting if less than such a Plain which shall be to the Rect-angle of the Legs as a Portion in the Base-line intercepted between two Lines from the Vertex making at the Base a like Angle with that of the Vertex to one of those two Lines so drawn VIII Of this we are to give great variety of Examples in the following Chapter where this General Theorem is applied to particular cases And which is further improved by these two ensuing Propositions IX The Radius of a Circle with the subtenses of two Arches being given the Subtense of their Aggregate is also given For supposing the subtenses of the given Arches to be A E The subtenses of their Remainders to a Semicircle are also had Suppose And And therefore inscribing a Quadrilater whose opposite sides are A ε and E α one of the Diagonals is the Diameter = 2R the other the Subtense of the Sum or Aggregate of those Arches suppose X. The same being given the Subtense of the Difference of those Arches is also given For having as before A α E ε 2R we have by a Quadrilater duly inscribed the Subtense of the Difference XI It is manifest also from what is before delivered that the same Triangle GΓμ doth indifferently serve for the Angle of 120 Degrees and of 60 Degrees And in like manner for 135 and 45 And so for any two Arches whereof one doth as much exceed as the other wants of a Quadrant For the Angle V is in both the same and the Angles at the Base differ only in this That in one the External Angle in the other the Internal which is the others Complement to two Right-angles is equal to the Angle of CD at the Vertex XII Hence it follows That Of two Angles where the Legs of the one are respectively equal to those of the other the one as much exceeding a Right-angle as the other wants of it The Square of the Base in the one doth as much exceed the two Squares of the Legs as in the other it wants thereof XIII And consequently In any Right-lined Triangle however inclined the Squares of the Axis or Diameter and of the half Bases twice taken are equal to the Squares of the Legs For supposing C C the two halfs of the Base and B the Diameter or Axis of the Triangle meaning thereby a streight Line from the Vertex to the middle of the Base and B β the two Legs It is manifest that of the two Angles at the Base which are each others Complement to two Right-angles the one doth as much exceed as the other wants of a Right-angle And therefore the Square of one of the Legs as Bq doth as much exceed as the other βq doth come short of Dq+Cq And therefore both together Bq+βq = 2Dq+2Cq XIV And therefore The Base and Axis or Diameter of a Triangle remaining the same however differently inclined the Aggregate of the Squares of the two Legs remains the same XV. And the same is to be understood of the Squares of Tangents of a Parabola Hyperbola Elipsis or other Curve Line having Diameter and Ordinates from the two ends of an Inscribed Ordinate to the Point of the Diameter produced if need be wherein those Tangents meet XVI The same may be likewise accommodated to the Segments of such Legs or Tangents Cut off by Lines Parallel to the Base Namely The Squares of such Segments intercepted by those Parallels together taken the Axe of such Trapezium remaining the same are the same Whether such Trapezium be Erect or however inclined For such Segments are still Proportional to their Wholes CHAP. VII Application thereof to particular cases I. IF A be a Right-angle or of 90 Degrees GΓ are Co-incident and μ = 0. and therefore And consequently by § 7 Chap. preced II. If A = 120 Degrees then is V that is the Angle contained of GΓ = 60 Degrees as being always the Difference of 2 A from two Right-angles And consequently GΓμ an Equilater Triangle for such also are the Angles at the Base each of which is the Complement of A to two Right-angles And therefore μ = G and Bq = Cq+Dq+CD III. If A = 60 Degrees Then also is V = 60 Degrees and μ = G as before And therefore Bq = Cq+Dq − CD IV. If A = 135. Then V = 90 And therefore by § 1. μq = Gq+Γq that is because G = Γ μq = 2Gq and And therefore V. If A = 45. Then also V = 90 And therefore as before and consequently VI. If A = 150 Then V = 120. And therefore by § 2. μq = Gq+Γq+GΓ that is because G = Γ μq = 3Gq and And VII If A = 30 Then V = 120. And therefore by § 2. μq = Gq+Γq+GΓ that is because G = Γ μq = 3Gq and And VIII If A = 157½ Then V = 135. And by § 4. And therefore IX If A = 22½ Then V = 135. And by § 4. And therefore X. If A = 112½ Then V = 45. And by § 5. And therefore XI If A = 6 − ½ Then V = 45. And by § 5. And therefore XII If A = 165 Then V = 150. And by § 6. And therefore XIII If A = 15 Then V = 150. And by § 6. And therefore XIV If A = 105 Then V = 30. And by § 7. And therefore XV. If A = 75 Then V = 30. And by § 7. And therefore XVI If A = 172½ Then V = 165. And by § 12. And therefore XVII If A = 7½ Then V = 165. And by § 12. And therefore XVIII If A = 97½ Then V = 15. And by § 13. And XIX If A = 82½ Then V = 15. And by § 13. And XX. If A = 142½ Then V = 105. And by § 14. And XXI If A = 37½ Then V = 105. And by § 14. And XXII If A = 127½ Then V = 75. And by § 15. And XXIII If A = 52½ Then V = 75. And by § 15. And And in
or E less than a Quadrant then A B and A D will be opposite sides and B C Diagonals And therefore CB − AB = AD. And consequently into equal to AD. That is And as before And for the same reason And LXXXVIII If the Subtense of the single Arch be P or S greater than a Quadrant and even greater than a Trient but less than two Trients Then B C and B P or B S will be opposite sides and D P or D S Diagonals And therefore BC+BP = PD or BC+BS = SD And consequently into equal to PD That is And As before And by the same reason BC+BS = SD if S also be greater than a Trient and LXXXIX But if the single Arch be that of S greater than a Quadrant but less than a Trient or P greater than two Trients but less than three Quadrants then B C and D S are opposite sides and B S Diagonals And therefore BS − BC = DS. And consequently into That is And As before And in like manner BP − BC = PD if the Arch of P be greater than two Trients which is the same as if less than one and XC From all which ariseth this General Theorem The Rect-angle of the Subtenses of the single and of the Quadruple Arch is equal to the Subtense of the double Multiplied into the Excess of the Subtense of the Triple above that of the single in case this be less than a Quadrant or more than three Quadrants or into the Excess of the Subtense of the single above that of the Triple in case the single be more than a Quadrant but less than a Trient or more than two Trients but less than three Quadrants or lastly into the Sum of the Subtenses of the Triple and single in case this be more than a Trient but less than two Trients That is AD = B into C − A if the Arch of A be less than a Quadrant or greater than three Quadrants A − C if it be greater than a Quadrant but less than a Trient or greater than two Trients but less than three Quadrants A+C if it be greater than a Trient but less than two Trients XCI And universally That is if the Difference of 2RqA and A c whereof that is the greater if the single Arch be less than a Quadrant or greater than three Quadrants but this if contrarywise divided by Rc be Multiplied into Product is equal to D. XCII And therefore That is XCIII As the Cube of the Radius to the Solid of the Subtense of the single Arch into the Difference of the Square of it self and of the double Square of the Radius So is the Subtense of the Difference of that single Arch from a Semicircumference to the Subtense of the Quadruple Arch. XCIV Now what was before said at § 15 Chap. 29. That the Subtense of an Arch with that of its Remainder to a Semicircumference or of its Excess above a Semicircumference will require the same Subtense of the double Arch is the same as to say that From any Point of Circumference two Subtenses drawn to the two ends of any inscribed Diameter as A E will require the same Subtense B of the double Arch. XCV And what is said at § 12 26 Chap. preced That the Subtense of an Arch less than a Trient and of its Residue to a Trient as A E and of a Trient increased by either of those as Z will have the same Subtense of the Triple Arch is the same in effect with this that From any Point of the Circumference three subtenses drawn to the three Angles of any inscribed Regular Trigone as A E Z will have the same Subtense C of the Triple Arch. XCVI And what is said here at § 18 20. That the Subtense of an Arch less than a Quadrant and of its Residue to a Quadrant as A E and of a Quadrant increased by either of these as P S will have the same Subtense of the Quadruple Arch Is the same with this that From any Point of the Circumference Four Subtenses drawn to the four Angles of any inscribed Regular Tetragone as A E P S will have the same Subtense D of the Quadruple Arch. XCVII But the same holds respectively in other Multiplications of Arches as five Subtenses from the same Point to the five Angles of an inscribed Regular Pentagon and six to the six Angles of an Hexagon c. Will have the same Subtense of the Arches Quintuple Sextuple c. For they all depend on the same common Principle That a Semicircumference Doubled a Trient Tripled a Quadrant Quadrupled a Quintant Quintupled a Sextant Sextupled c. Make one entire Revolution which as to this business is the same as nothing And therefore universally XCVIII From any Point of the Circumference two three four five six or more subtenses drawn to so many ends of the Diameter or Angles of a Regular Polygone of so many Angles however inscribed will have the same Subtense of the Arch Multiplied by the number of such ends or Angles And therefore CXIX An Equation belonging to such Multiplication or Section of an Arch or Angle must have so many Roots Affirmative or Negative as is the Exponent of such Multiplication or Section As two for the Bisection three for the Trisection four for the Quadrisection five for the Quinquisection And so forth C. And consequently Such Equations may accordingly be resolved by such Section of an Angle As was before noted at § 61 Chap. preced of the Trisection of an Angle CHAP. IV. Of the Quintuplation and Quinquisection of an ARCH or ANGLE I. IF in a Circle be inscribed a Quadrilater whose sides A F the Subtenses of the single Arch and the Quintuple be Parallel B B subtenses of the double opposite The Diagonals will be C C the subtenses of the Triple as is evident from the Figure But it is evident also that in this case the single Arch must be less than a Quintant or fifth part of the whole Circumference II. And therefore the Rect-angle of the Diagonals being equal to the two Rect-angles of the opposite sides Cq − Bq = AF. And by the same reasons cq − bq = EF. That is III. The Square of the Subtense of the Triple Arch wanting the Square of the Subtense of the double Arch is equal to the Rect-angle of the Subtenses of the single and of the Quintuple the single Arch being less than a fifth part of the whole Circumference IV. And therefore if it be divided by one of them it gives the other That is and And in like manner and V. But C+B into C − B is equal to Cq − Bq. And therefore That is VI. As the Subtense of the single Arch less than a fifth part of the whole Circumference to the Aggregate of the subtenses of the Triple and double so is the Excess of the Subtense of the Triple above that of the double to that of